Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 50

Find the equation of the tangent line of the curve $\displaystyle y = x^4 + 2x^2 - x$ at Point $(1,2)$

Required:

Equation of the tangent line to the curve at $P(1,2)$

Solution:

Let $y' = m$ (slope)


$
\begin{equation}
\begin{aligned}

\qquad y' = m =& \frac{d}{dx} (x^4) + \frac{d}{dx} (2x^2) - \frac{d}{dx} (x)
&& \text{}
\\
\\
\qquad y' = m =& 4x^3 + 4x - 1
&& \text{}
\\
\\
\qquad m =& 4x^3 + 4x - 1
&& \text{Substitute value of $x$ which is 1}
\\
\\
\qquad m =& 4(1)^3 + 4 (1) - 1
&& \text{}
\\
\\
m =& 7
&&

\end{aligned}
\end{equation}
$


Solving for the equation of the tangent line:


$
\begin{equation}
\begin{aligned}

\qquad y - y_1 =& m(x - x_1)
&& \text{Substitute the value of the slope $(m)$ and the given point}
\\
\\
\qquad y - 2 =& 7 (x -1)
&& \text{Distribute 7 to the equation}
\\
\\
\qquad y - 2 =& 7x - 7
&& \text{Add 2 to each side}
\\
\\
\qquad y =& 7x - 7 + 2
&& \text{Combine like terms}
\\
\\
\qquad y =& 7x - 5
&& \text{Equation of the tangent line to the curve at $P(1, 2)$}

\end{aligned}
\end{equation}
$